An Electron Falls in a Field

solved ElectricField · NewtonsSecondLaw

Q. An electron falls from rest $v_{0} = 0$ m/s, $x_{0} = 0$ m through a distance of 1.5 cm $x_{1} = 1.5$ cm in a uniform electric field $F_{net} = F$ of magnitude 2.0 × 10⁴ N/C $E = 2.0 \times 10^{4}$ N/C. Taking its charge as 1.6 × 10⁻¹⁹ C $q = 1.6 \times 10^{−19}$ C and its mass as 9.11 × 10⁻³¹ kg $m = 9.11 \times 10^{−31}$ kg, what is the time of fall, in seconds?

Inputs

distance

x_{1}

1.5 cm

field

E

2e+04 N/C

charge

q

1.6e-19 C

mass

m

9.11e-31 kg

Outputs

time

Δ t

How it's solved

Solved with ElectricField, NewtonsSecondLaw

Given

$E = 2.0 \times 10^{4}$ N/C
$q = 1.6 \times 10^{−19}$ C
$m = 9.11 \times 10^{−31}$ kg
$x_{1} = 1.5$ cm
$v_{0} = 0$ m/s from “falls from rest”
$x_{0} = 0$ m from “falls from rest”
$F_{net} = F$ from “in a uniform electric field”

Find $Δ t$ .

Working

$E = 2.0 \times 10^{4}$ N/C $= 20000$ V/m
$x_{1} = 1.5$ cm $= 0.015$ m
force on a charge in a field(electron) $F = q \cdot E$
$F = 1.6 \times 10^{−19} \cdot 20000$
$F = 3.2 \times 10^{−15}$ N
in a uniform electric field $F_{net} = F$
$F_{net} = 3.2 \times 10^{−15}$
$F_{net} = 3.2 \times 10^{−15}$ N
second law(falling) $F_{net} = m \cdot a$
$3.2 \times 10^{−15} = 9.11 \times 10^{−31} \cdot a$
$a = 3.51262349 \times 10^{15}$ m/s²
position after a time(falling) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$0.015 = 0 + 0 \cdot Δ t + \frac{3.51262349 \times 10^{15} \cdot {Δ t}^{2}}{2}$
$Δ t = 2.922 \times 10^{−9}$ s
$Δ t \approx 2.9 \times 10^{−9}$ sto 2 significant figures

Conclusion

Unlike a fall under gravity, the field's pull grows with the charge but not with the mass, so the light electron is flung through the gap in nanoseconds.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :an_electron_falls_in_a_field do4  kind :numerical5  describes "An electron falls from rest through a distance of %{distance} in a uniform electric field of magnitude " \6            "%{field}. Taking its charge as %{charge} and its mass as %{mass}, %{time}, in seconds?"7  given distance: "1.5 cm"8  given field: "2.0e4 N/C"9  given charge: "1.6e-19 C"10  given mass: "9.11e-31 kg"11  asks time: "what is the time of fall"12end

solution.rb

1require_relative "../questions/an_electron_falls_in_a_field"2require_relative "../science/electric_charges_and_fields/electric_field"3require_relative "../science/laws_of_motion/newtons_second_law"45solve :an_electron_falls_in_a_field, using: [ ElectricField, NewtonsSecondLaw ] do6  objects :electron7  electron.goes(:falling)89  take field: E10  take charge: electron.charge11  take mass: electron.mass12  take distance: electron.final_position1314  deduce_from "falls from rest", electron.initial_velocity => "0 m/s", electron.initial_position => "0 m"15  deduce_from "in a uniform electric field", electron.net_force => electron.force_of_the_field1617  take time: falling.time_taken18  answer find_value(:time, in: "s", figures: 2)1920  conclusion "Unlike a fall under gravity, the field's pull grows with the charge but not with the mass, so " \21             "the light electron is flung through the gap in nanoseconds."22end

playground.rb

1require_relative "../solutions/an_electron_falls_in_a_field"23play_with :an_electron_falls_in_a_field do4  title "An Electron Falls in a Field"5end

The laws it drew on

quantity	written	units	within	from
charge	$q$	C	—	ElectricField
coulomb constant	$k$	N m² C⁻²	—	ElectricField
field strength	$E$	V/m	—	ElectricField
separation	$r$	m	0 … ∞	ElectricField
field of a charge	$E_{q}$	V/m	—	ElectricField
force of the field	$F$	N	—	ElectricField
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$work$	—	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
displacement of electron during falling	$Δ x$	m	—
position of electron at end of falling	$x_{1}$	m	—
position of electron at start of falling	$x_{0}$	m	—
average velocity of electron during falling	$v_{av}$	m/s	—
time taken of falling	$Δ t$	s	0 … 1e+05
average speed of electron during falling	$c_{av}$	m/s	—
path length of electron during falling	$d$	m	—
change of velocity of electron during falling	$Δ v$	m/s	—
velocity of electron at end of falling from ground	$v_{1}$	m/s	—
velocity of electron at start of falling from ground	$v_{0}$	m/s	—
average acceleration of electron during falling	$a_{av}$	m/s²	—
acceleration of electron during falling	$a$	m/s²	—
net force on electron during falling	$F_{net}$	N	—
mass of electron	$m$	kg	0 … ∞
force of the field on ground	$F_{g}$	N	—
charge of ground	$q_{g}$	C	—
force of the field on electron	$F$	N	—
charge of electron	$q$	C	—
field of a charge of ground by electron	$E_{q,g}$	V/m	—
separation of ground from electron	$r_{g}$	m	0 … ∞
field of a charge of electron by ground	$E_{q,g}$	V/m	—
separation of electron from ground	$r$	m	0 … ∞
weight on ground	$W_{g}$	N	—
weight on electron	$W$	N	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

F_{b} = q_{b} \cdot E

force on a charge in a field

E_{q,ps} = \frac{k \cdot q_{s}}{{r_{ps}}^{2}}

field of a point charge

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(falling)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(falling)

c_{av} = \frac{d}{Δ t}

path a time(falling)

Δ v = v_{1} - v_{0}

velocity gained(falling)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(falling)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(falling)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(falling)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(falling)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(falling)

F_{net} = m \cdot a

second law(falling)

F_{g} = q_{g} \cdot E

force on a charge in a field(ground)

F = q \cdot E

force on a charge in a field(electron)

E_{q,g} = \frac{k \cdot q}{{r_{g}}^{2}}

field of a point charge(ground, electron)

E_{q,g} = \frac{k \cdot q_{g}}{r^{2}}

field of a point charge(electron, ground)

F_{net} = F

in a uniform electric field

m \neq 0

has mass?(electron)

F_{net} = 0

free of forces?(electron, falling)